<meta http-equiv="refresh" content="1; url=/nojavascript/"> Quadratic Formula ( Read ) | Algebra | CK-12 Foundation

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What if you needed to solve the quadratic equation $x^2+x-5=0$ in order to determine the width of a rectangular map? You graphed the function $f(x)=x^2+x-5$ , but from the graph you can only tell the approximate width, and you want a more precise answer. In this Concept, you'll learn to use the quadratic formula to solve quadratic equations like the one representing this situation so that you can get exact solutions to the equations.

### Guidance

Previous Concepts have presented three methods to solve a quadratic equation:

• By graphing to find the zeros;
• By solving using square roots; and
• By using completing the square to find the solutions

This Concept will present a fourth way to solve a quadratic equation: using the quadratic formula.

As early as 1200 BC, people were interested in solving quadratic equations. The Babylonians solved simultaneous equations involving quadratics. In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit formula to solve a quadratic equation. The quadratic formula was written as it is today by the Arabic mathematician Al-Khwarizmi. It is his name upon which the word “Algebra” is based.

The solution to any quadratic equation in standard form, $0=ax^2+bx+c$ , is:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

#### Example A

Solve $x^2+10x+9=0$ using the quadratic formula.

Solution:

We know from the last Concept the answers are $x=-1$ or $x=-9$ .

By applying the quadratic formula and $a=1, b=10$ , and $c=9$ , we get:

$x &= \frac{-10 \pm \sqrt{(10)^2-4(1)(9)}}{2(1)}\\x &= \frac{-10 \pm \sqrt{100-36}}{2}\\x &= \frac{-10 \pm \sqrt{64}}{2}\\x &= \frac{-10 \pm 8}{2}\\x &= \frac{-10 + 8}{2} \ or \ x=\frac{-10-8}{2}\\x &= -1 \ or \ x=-9$

#### Example B

Solve $-4x^2+x+1=0$ using the quadratic formula.

Solution:

$\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\text{Plug in the values} \ a=-4, b=1, c=1: && x& =\frac{-1 \pm \sqrt{(1)^2-4(-4)(1)}}{2(-4)}\\\text{Simplify:} && x & =\frac{-1 \pm \sqrt{1+16}}{-8}=\frac{-1 \pm \sqrt{17}}{-8}\\\text{Separate the two options:} && x&=\frac{-1+\sqrt{17}}{-8} \ \text{and} \ x=\frac{-1-\sqrt{17}}{-8}\\\text{Solve:} && x & \approx -.39 \ \text{and} \ x \approx .64$

#### Example C

Solve $8t^2+10t+3=0$ using the quadratic formula.

Solution:

$\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\text{Plug in the values} \ a=8, b=10, c=3: && x& =\frac{-10 \pm \sqrt{(10)^2-4(8)(3)}}{2(8)}\\\text{Simplify:} && x & =\frac{-10 \pm \sqrt{100+96}}{16}=\frac{-10 \pm \sqrt{196}}{16}\\\text{Separate the two options:} && x&=\frac{-10+\sqrt{196}}{16} \ \text{and} \ x=\frac{-10-\sqrt{196}}{16}\\\text{Separate the two options:} && x&=\frac{-10+14}{16} \ \text{and} \ x=\frac{-10-14}{16}\\\text{Solve:} && x & =\frac{1}{4} \ \text{and} \ x =-\frac{3}{2}$

### Guided Practice

Solve $3k^2+11k=4$ using the quadratic formula.

Solution:

First, we must make it so one side is equal to zero:

$3k^2+11k=4 \Rightarrow 3k^2+11k-4=0$

Now it is in the correct form for using the quadratic formula.

$\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\text{Plug in the values} \ a=3, b=11, c=-4: && x& =\frac{-11 \pm \sqrt{(11)^2-4(3)(-4)}}{2(3)}\\\text{Simplify:} && x & =\frac{-11 \pm \sqrt{121+48}}{6}=\frac{-11 \pm \sqrt{169}}{6}\\\text{Separate the two options:} && x&=\frac{-11+\sqrt{169}}{6} \ \text{and} \ x=\frac{-11-\sqrt{169}}{6}\\\text{Separate the two options:} && x&=\frac{-11+13}{6} \ \text{and} \ x=\frac{-11-13}{6}\\\text{Solve:} && x & =\frac{1}{3} \ \text{and} \ x =-4$

1. What is the quadratic formula? When is the most appropriate situation to use this formula?
2. When was the first known solution of a quadratic equation recorded?

1. $x^2+4x-21=0$
2. $x^2-6x=12$
3. $3x^2-\frac{1}{2}x=\frac{3}{8}$
4. $2x^2+x-3=0$
5. $-x^2-7x+12=0$
6. $-3x^2+5x=0$
7. $4x^2=0$
8. $x^2+2x+6=0$

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